A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2– is colourless. Explain.
In case of [Ni(H2O)6]2+ H2O is a weak field ligand , so it does not cause the pairing of the unpaired electron of Ni2+ ion. Thus there is possibility of the intra d-d transition from the d orbital of lower energy to that of higher energy. Thus the light is absorbed from the visible region and complimentary colour is observed. But in case of [Ni(CN)4]2– CN- is strong field ligand.
Therefore it will cause pairing of the unpaired electrons of Ni2+ ion. There are no unpaired electrons present, so there is no d-d transition and hence it is colourless.
Using IUPAC norms write the formulas for the following:
(i) Tetrahydroxidozincate (II)
(ii) Potassium tetrachloridopalladate (II)
(iii) Diamminedichloridoplatinum (II)
(iv) Potassium tetracyanidonickelate (II)
(v) Pentaamminenitrito-O-cobalt (III)
(vi) Hexaamminecobalt (III) sulphate
(vii) Potassium tri (oxalato) chromate (III)
(viii) Hexaammineplatinum (IV)
(ix) Tetrabromidocuprate (II)
(x) Pentaamminenitrito-N-cobalt (III)
Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:
(i) K[Cr(H2O)2(C2O4)2].3H2O (iii) [CrCl3(py)3] (v) K4[Mn(CN)6]
(ii) [Co(NH3)5Cl-]Cl2 (iv) Cs[FeCl4]